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Math Help - Probablility help with combination/permutations

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    Probablility help with combination/permutations

    Solve for n if nP5 = 120 x nC3
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  2. #2
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    Quote Originally Posted by mathamatics112 View Post
    Solve for n if nP5 = 120 x nC3
    <br />
^nP_r = \frac{n!}{(n-r)!}

    ^nC_r = \frac{n!}{(n-r)!r!}


    ^nP_5 = \frac{n!}{(n-5)!} = n(n-1)(n-2)(n-3)(n-4) (eq1)

    120\cdot ^nC_3 = 120\cdot\frac{n!}{(n-3)!3!} = 120\cdot \frac{n(n-1)(n-2)}{6} (eq2)

    eq1 = eq2

    n(n-1)(n-2)(n-3)(n-4) = 20(n)(n-1)(n-2)

    n(n-1)(n-2) will cancel:

    n^2-7n-8 = 0

    solve to get n = 8 or n = -1.

    However since n must be greater than 0 only n = 8 is a solution
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